Count subarrays made up of single-digit integers only

Given an array arr[] consisting of N positive integers, the task is to count subarrays consisting of single-digit elements only.

Examples:

Input: arr[] = {0, 1, 14, 2, 5}
Output: 6
Explanation: All subarrays made of only single digit numbers are {{0}, {1}, {2}, {5}, {0, 1}, {2, 5}}. Therefore, the total count of subarrays is 6.

Input: arr[] ={12, 5, 14, 17}
Output: 1
Explanation: All subarrays made of only single digit numbers are {5}.
Therefore, the total count of subarrays is 1.

Naive Approach: The simplest approach is to traverse the array and generate all possible subarrays. For each subarray, check if all integers in it are single-digit integers or not. 

Time Complexity: O(N³)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to find the size of each block of contiguous single-digit integers and increment the count by total subarrays of that length. Follow the steps below to solve the problem:

• Initialize a variable, say res = 0 and c = 0, to store the total count of subarrays and the total count of single-digit integers in a subarray.
• Traverse the array and perform the following operations:
  • If arr[i] < 10, increment the count of c by one and count of res by c.
  • Otherwise, assign c = 0.
• Finally, print the total count of single-digit integer subarrays.

Below is the implementation of the above approach:

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to count of subarrays made
// up of single digit integers only
int singleDigitSubarrayCount(int arr[],
                             int N)
{
    // Stores count of subarrays
    int res = 0;

    // Stores the count of consecutive
    // single digit numbers in the array
    int count = 0;

    // Traverse the array
    for (int i = 0; i < N; i++) {

        if (arr[i] <= 9) {

            // Increment size of block by 1
            count++;

            // Increment res by count
            res += count;
        }

        else {

            // Assign count = 0
            count = 0;
        }
    }

    cout << res;
}

// Driver Code
int main()
{
    // Given array
    int arr[] = { 0, 1, 14, 2, 5 };

    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
    singleDigitSubarrayCount(arr, N);

    return 0;
}

// Java program for the above approach
class GFG
{

// Function to count of subarrays made
// up of single digit integers only
static void singleDigitSubarrayCount(int arr[],
                             int N)
{
  
    // Stores count of subarrays
    int res = 0;

    // Stores the count of consecutive
    // single digit numbers in the array
    int count = 0;

    // Traverse the array
    for (int i = 0; i < N; i++) 
    {
        if (arr[i] <= 9)
        {

            // Increment size of block by 1
            count++;

            // Increment res by count
            res += count;
        }

        else 
        {

            // Assign count = 0
            count = 0;
        }
    }
    System.out.print(res);
}

// Driver Code
public static void main(String[] args)
{
  
    // Given array
    int arr[] = { 0, 1, 14, 2, 5 };

    // Size of the array
    int N = arr.length;
    singleDigitSubarrayCount(arr, N);
}
}

// This code is contributed by 29AjayKumar 

# Python3 program for the above approach

# Function to count of subarrays made
# up of single digit integers only
def singleDigitSubarrayCount(arr, N):
    
    # Stores count of subarrays
    res = 0

    # Stores the count of consecutive
    # single digit numbers in the array
    count = 0

    # Traverse the array
    for i in range(N):
        if (arr[i] <= 9):

            # Increment size of block by 1
            count += 1

            # Increment res by count
            res += count
        else:
            # Assign count = 0
            count = 0
    print (res)

# Driver Code
if __name__ == '__main__':
  
    # Given array
    arr = [0, 1, 14, 2, 5]

    # Size of the array
    N = len(arr)
    singleDigitSubarrayCount(arr, N)

    # This code is contributed by mohit kumar 29.

// C# program for the above approach
using System;
class GFG{

// Function to count of subarrays made
// up of single digit integers only
static void singleDigitSubarrayCount(int[] arr,
                             int N)
{
  
    // Stores count of subarrays
    int res = 0;

    // Stores the count of consecutive
    // single digit numbers in the array
    int count = 0;

    // Traverse the array
    for (int i = 0; i < N; i++) 
    {
        if (arr[i] <= 9)
        {

            // Increment size of block by 1
            count++;

            // Increment res by count
            res += count;
        }
        else 
        {

            // Assign count = 0
            count = 0;
        }
    }
    Console.Write(res);
}

// Driver Code
public static void Main(string[] args)
{
    // Given array
    int[] arr = { 0, 1, 14, 2, 5 };

    // Size of the array
    int N = arr.Length;
    singleDigitSubarrayCount(arr, N);
}
}

// This code is contributed by sanjoy_62.

<script>

// Javascript program for the above approach

// Function to count of subarrays made
// up of single digit integers only
function singleDigitSubarrayCount(arr, N)
{
    
    // Stores count of subarrays
    let res = 0;

    // Stores the count of consecutive
    // single digit numbers in the array
    let count = 0;

    // Traverse the array
    for(let i = 0; i < N; i++) 
    {
        if (arr[i] <= 9)
        {
            
            // Increment size of block by 1
            count++;

            // Increment res by count
            res += count;
        }
        else 
        {
            
            // Assign count = 0
            count = 0;
        }
    }
    document.write(res);
}

// Driver Code

// Given array
let arr = [ 0, 1, 14, 2, 5 ];

// Size of the array
let N = arr.length;

singleDigitSubarrayCount(arr, N);

// This code is contributed by Manoj.

</script>

6

Time Complexity: O(N)
Auxiliary Space: O(1) 

Approach#2: Using nested loop

This approach is  a brute-force solution. The code loops over all possible subarrays of the input array and checks if each subarray is made up of single-digit integers only. If it is, the count of such subarrays is incremented.

Algorithm

1. Initialize a variable count to 0.
2. Loop through the array using two nested loops. The outer loop iterates from 0 to n-1, where n is the length of the input array. The inner loop iterates from the current outer loop index to n-1.
3. For each subarray formed by the outer and inner loop indices, check if it contains only single-digit integers. This is done using a nested loop that iterates through each element of the subarray and checks if it is a single-digit integer.
4. If the subarray is made up of only single-digit integers, increment the count variable.
5. After looping through all subarrays, return the count.

#include <iostream>
#include <vector>

using namespace std;

int count_single_digit_subarrays(const vector<int>& arr) {
    int n = arr.size();
    int count = 0;
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            bool is_single_digit_subarray = true;
            for (int k = i; k <= j; k++) {
                if (arr[k] < 0 || arr[k] > 9) {
                    is_single_digit_subarray = false;
                    break;
                }
            }
            if (is_single_digit_subarray) {
                count++;
            }
        }
    }
    return count;
}

int main() {
    vector<int> arr1 = {0, 1, 14, 2, 5};
    cout << count_single_digit_subarrays(arr1) << endl;

    vector<int> arr2 = {12, 5, 14, 17};
    cout << count_single_digit_subarrays(arr2) << endl;

    return 0;
}

import java.util.*;

public class GFG {
    // Function to count the number of single-digit subarrays in the given vector
    public static int countSingleDigitSubarrays(List<Integer> arr) {
        int n = arr.size();
        int count = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                boolean isSingleDigitSubarray = true;
                for (int k = i; k <= j; k++) {
                    if (arr.get(k) < 0 || arr.get(k) > 9) {
                        isSingleDigitSubarray = false;
                        break;
                    }
                }
                if (isSingleDigitSubarray) {
                    count++;
                }
            }
        }
        return count;
    }

    // Driver Code
    public static void main(String[] args) {
        List<Integer> arr1 = Arrays.asList(0, 1, 14, 2, 5);
        System.out.println(countSingleDigitSubarrays(arr1));

        List<Integer> arr2 = Arrays.asList(12, 5, 14, 17);
        System.out.println(countSingleDigitSubarrays(arr2));
    }
}

def count_single_digit_subarrays(arr):
    n = len(arr)
    count = 0
    for i in range(n):
        for j in range(i, n):
            is_single_digit_subarray = True
            for k in range(i, j+1):
                if arr[k] < 0 or arr[k] > 9:
                    is_single_digit_subarray = False
                    break
            if is_single_digit_subarray:
                count += 1
    return count

arr1 = [0, 1, 14, 2, 5]
print(count_single_digit_subarrays(arr1))  

arr2 = [12, 5, 14, 17]
print(count_single_digit_subarrays(arr2)) 

using System;

class Program
{
    // Function to count the number of subarrays with single-digit elements
    static int CountSingleDigitSubarrays(int[] arr)
    {
        int n = arr.Length;
        int count = 0;

        // Loop through all possible subarrays
        for (int i = 0; i < n; i++)
        {
            for (int j = i; j < n; j++)
            {
                bool isSingleDigitSubarray = true;

                // Check if each element in the subarray is a single digit
                for (int k = i; k <= j; k++)
                {
                    if (arr[k] < 0 || arr[k] > 9)
                    {
                        isSingleDigitSubarray = false;
                        break;
                    }
                }

                // If the subarray consists of single-digit elements, increment the count
                if (isSingleDigitSubarray)
                {
                    count++;
                }
            }
        }

        return count;
    }

    static void Main()
    {
        // Example 1
        int[] arr1 = { 0, 1, 14, 2, 5 };
        Console.WriteLine(CountSingleDigitSubarrays(arr1));

        // Example 2
        int[] arr2 = { 12, 5, 14, 17 };
        Console.WriteLine(CountSingleDigitSubarrays(arr2));
    }
}

function countSingleDigitSubarrays(arr) {
    const n = arr.length;
    let count = 0;

    // Loop through the array elements
    for (let i = 0; i < n; i++) {
        for (let j = i; j < n; j++) {
            let isSingleDigitSubarray = true;

            // Check if each element within the subarray is a single digit
            for (let k = i; k <= j; k++) {
                if (arr[k] < 0 || arr[k] > 9) {
                    isSingleDigitSubarray = false;
                    break;
                }
            }

            // If the subarray contains only single-digit elements, increment count
            if (isSingleDigitSubarray) {
                count++;
            }
        }
    }

    return count;
}

const arr1 = [0, 1, 14, 2, 5];
console.log(countSingleDigitSubarrays(arr1)); // Output: 9

const arr2 = [12, 5, 14, 17];
console.log(countSingleDigitSubarrays(arr2)); // Output: 4

6
1

Time Complexity: O(n^3), where n is the length of the input array. This is because the code uses three nested loops to iterate through all possible subarrays of the input array.

Auxiliary Space: O(1), as it only uses a constant amount of extra space to store the count variable. The input array is not modified, and no extra data structures are used.